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I have to compute the infinity laplacian $$ \Delta_{\infty}u=\sum_{i,j=1}^n\frac{\partial^2u}{\partial x_ix_j}\frac{\partial u}{\partial x_i}\frac{\partial u}{\partial x_j} $$ of a radial function $R(r)=R(|x|)$. The answer is $\Delta_{\infty}R(r)=R''(r)R'(r)^2$. My attempt is the following. We have $$ \frac{\partial R}{\partial x_i}=R'\frac{x_i}{r},\qquad \frac{\partial R}{\partial x_j}=R'\frac{x_j}{r} $$ and $$ \frac{\partial^2R}{\partial x_ix_j}=\frac{x_ix_j}{r^2}\left(R''-\frac{R'}{r}\right). $$ So $$ \Delta_{\infty}R(r)=\sum_{i,j=1}^n\frac{x_ix_j}{r^2}\left(R''-\frac{R'}{r}\right)\frac{x_ix_j}{r^2}R'^2=\left(R''-\frac{R'}{r}\right)\frac{R'^2}{r^4}\sum_{i,j=1}^nx_i^2x_j^2=\left(R''-\frac{R'}{r}\right)R'^2 $$ but this does not coincide with the answer due to the term $\frac{R'}{r}$. I cannot say the error. Can someone help me?

Thank you

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The mistake comes from forgetting the case when $i = j $ in the double derivative. The actual formula is: $$ \frac{\partial^2 R}{\partial x_i x_j} = \frac{x_i x_j}{r^2} \left( R^{\prime \prime} - \frac{R^\prime} {r} \right) + \delta_j(i) \frac{R^\prime}{r}$$

Therefore the infinite laplacian is:

$$ \Delta_{\infty}R(r)=\sum_{i,j=1}^n\left( \frac{x_ix_j}{r^2}\left(R''-\frac{R'}{r}\right) + \delta_j(i) \frac{R^\prime}{r} \right) \frac{x_ix_j}{r^2}R'^2 $$

Using your same computations, we have that $\begin{align} \Delta_{\infty}R(r) & = \left( R^{\prime \prime} - \frac{R^\prime}{r}\right) R^{\prime 2} + \sum_{i,j= 1}^n \delta_j(i) \frac{R^\prime}{r} \frac{x_i x_j}{r^2} R^{\prime 2} \\ & = \left( R^{\prime \prime} - \frac{R^\prime}{r}\right) R^{\prime 2} + \sum_{i=1}^n \frac{R^\prime}{r} \frac{x_i^2}{r^2} R^{\prime 2} \\ & = \left( R^{\prime \prime} - \frac{R^\prime}{r}\right) R^{\prime 2} + \frac{R^\prime}{r}R^{\prime 2} \\ & = R^{\prime \prime} R^{\prime 2}. \end{align}$

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  • $\begingroup$ In the derivative $\frac{\partial^2R}{\partial x_ix_j}=\frac{x_ix_j}{r^2}\left(R''-R'\right)+\delta_j(i)\frac{R'}{r}$ shall we have the term $\frac{R'}{r}$ in $(R''-R')$ in place of $R'$, right? $\endgroup$ – Redeldio Nov 24 '17 at 3:30
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    $\begingroup$ Yes! I fix it right away. $\endgroup$ – pancho Nov 24 '17 at 8:34

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